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February 5[edit]

maximization[edit]

How do I do this maximization problem?

$\max _{c_{i}}{\frac {1}{n}}\sum _{i=1}^{n}\log(c_{i})$

subject to

$\sum _{i=1}^{n}p_{i}c_{i}=h$

Cannot use first order conditions here. But because it is maximizing over an average of a sum of logarithms, I can see that all the $c_{i}$ should be equal. The answer is $c_{i}=h/(np_{i})$ but I am unsure how to verify this. —Preceding unsigned comment added by 130.102.158.15 (talk) 01:32, 5 February 2011 (UTC)[reply]

Seems like a pretty straightforward application of Lagrange multipliers. I'd write more, but the article explains it better. Invrnc (talk) 04:04, 5 February 2011 (UTC)[reply]

The Arithmetic-geometric mean inequality shows that you have achieved the maximum. Sławomir Biały (talk) 13:23, 5 February 2011 (UTC)[reply]

gradient of scalar product[edit]

hi, any help with proving that grad (v_ . r_) = v_ using spherical polars, where v_ is a uniform vector field would be great it is trivial to prove using summation convention or Cartesian coordinates but having to use spherical polars looks messy... v_.r_ alone gives lots of terms containing sines and cosines of thetas and phis... i have read the wikipage on it, but perhaps i have not fully grasped the way it works or what the question is after... —Preceding unsigned comment added by 131.111.222.12 (talk) 02:17, 5 February 2011 (UTC)[reply]

Since it is indeed easy in Cartesian coordinates, what makes you want to do it in spherical? This looks like a prime candidate for the "convert to a frame where things are easy, then convert back again if necessary" strategy.

If this is homework, the point of the exercise may be to allow you to see for yourself how most of the horrible trigonometric terms cancel each other out once you differentiate. So carry on, and if you actually get stuck instead of just losing courage, then show some work here. –Henning Makholm (talk) 09:39, 5 February 2011 (UTC)[reply]

thanks, henning. i have proved it. —Preceding unsigned comment added by 131.111.222.12 (talk) 17:30, 5 February 2011 (UTC)[reply]

Multiplicative inverse function?[edit]

Hi. This is not a homework question. What is the term used to describe the following type of math done to a quadratic function: $nx^{n-1}-mx^{m-1}$ ? I think it may be a "multiplicative inverse", but this is done to find the highest possible value of the area, and what is a simplification thereof? Thanks. ~A H 1^(TCU) 02:39, 5 February 2011 (UTC)[reply]

Based on the context of maximizing, you might be thinking of the derivative, i.e., the derivative of $x^{n}-x^{m}$ with respect to x is $nx^{n-1}-mx^{m-1}$ . --Kinu ^t/_c 02:48, 5 February 2011 (UTC)[reply]

A calculus problem[edit]

Dear Wikipedians:

I am working on the a problem that seems to employ the fundamental theorem of calculus. The problem asks me to show that

$\int _{0}^{x}f(u)(x-u)du=\int _{0}^{x}\left(\int _{0}^{u}f(t)dt\right)du$

given a continuous function f.

My basic approach is to let $G(x)=\int _{0}^{x}f(u)(x-u)du$ and $F(x)=\int _{0}^{x}\left(\int _{0}^{u}f(t)dt\right)du$ and to show that $G''(x)=F''(x)$ .

For $F(x)$ , we have:

$F'(x)=\int _{0}^{x}f(t)dt$

$F''(x)=f(x)$

However, for $G(x)$ I am encountering great difficulty in finding its second derivative. How do I show that $G''(x)=f(x)$ ?

Thanks,

70.29.26.56 (talk) 05:30, 5 February 2011 (UTC)[reply]

Let's try something:

\int _{0}^{x}f(u)(x-u)\,du=\int _{0}^{x}f(u)\left(\int _{u}^{x}\,dt\right)\,du=\int _{0}^{x}\left(\int _{u}^{x}f(u)\,dt\right)\,du

The last step relies on the fact that ƒ(u) does not change as t goes from u to x, so ƒ(u) is a "constant" as a function of t. Then we can reverse the order of integration if we're careful about the bounds:

\int _{0}^{x}\int _{0}^{t}f(u)\,du\,dt.

And that's the same thing as what you've got. Michael Hardy (talk) 07:11, 5 February 2011 (UTC)[reply]

Thanks Michael! Your solution is so beautiful! 174.88.241.210 (talk) 16:13, 5 February 2011 (UTC)[reply]

Resolved

Primitive Roots[edit]

Just learning about basic number theory and primitive roots and I wrote a program to find a first few primes that have two/three as a primitive root. But when I started for primes which have 4 as a primitive root, I can't find any solution. I tried up to some large numbers but couldn't find a single solution. So my question is, is it possible for a prime to have 4 as a primitive root? How can we do this analytically? How can we find a prime for which 4 is a primitive root? Or if there doesn't exist one, then how would I go about proving it? Thanks!174.29.78.245 (talk) 06:25, 5 February 2011 (UTC)[reply]

Suppose 4 is a primitive root. Then 2 is a primitive root (if

4^{m}=x

, then

2^{2m}=x

). So 2 has order

p-1

. But

p-1

is even, so

4=2^{2}

has order

(p-1)/2

, contradicting its being a primitive root. So there are no such primes.--203.97.79.114 (talk) 08:07, 5 February 2011 (UTC)[reply]

proper class in ZFC[edit]

Let r be some arbitrary real number, which might be undefinable.

Is "all elements of V except for r" a proper class? That is, are there uncountably many proper classes? One could even say "arbitrary ordinal" instead of "arbitrary real", so there would be more classes than sets.

The identification of classes with formulas doesn't seem all that satisfying. 71.141.88.54 (talk) 18:39, 5 February 2011 (UTC)[reply]

Yes, V\{r} has to be a proper class. It cannot be a set, because the union of two sets is a set, and V=(V\{r})u{r} is definitely not a set.

Strictly speaking, ZFC does not allow you to even speak about proper classes in formal statements, so "uncountably many proper classes" is not a meaningful concept. NBG set theory is a conservative extension of ZFC in which one can speak about (and quantify over) proper classes, but even there it is unclear which formal sense "uncountably many proper classes" might have.

If we ask for a bijection between proper classes and the integers, it is clear that such a bijection cannot exist inside the theory, because it would need to contain proper classes as elements, and nothing can do that. In NBG one could ask whether there is a logical formula

\phi (n,X)

that bijectively assigns a proper class X to each integer n and vice versa. I don't think such a formula can possibly exist, but I cannot prove it offhand. (It's certainly impossible in Morse-Kelley set theory, but that might conceivably be because MK has more proper classes than NBG).

On the other hand, by the Skolem-Löwenheim theorem, if NBG is consistent, then it has a countable model, and in this model there would of course be meta-countably many proper classes -- see Skolem's paradox. –Henning Makholm (talk) 20:42, 5 February 2011 (UTC)[reply]

Thanks. I was thinking about the ZFC-centric view that classes are not formalized in ZFC itself, but are defined in terms of formulas in the metalanguage. But I guess that means classes are relative to specific models; for undefinable r, there has to be a constant symbol for r in the metalanguage before we can say "V\{r}". \ is a set operation anyway, so we can only use it in the context where V denotes the elements of a model, if I have it right. The apparent contradiction between "classes are denoted by formulas" and "uncountably many classes" is resolved by the idea that since the metalanguage can have uncountably many constant symbols, it can have uncountably many formulas. 71.141.88.54 (talk) 23:59, 5 February 2011 (UTC)[reply]

Normally, when you talk about definable proper classes, you mean definable with parameters. If you like, sure, you can add a constant symbol to represent the parameter, but it's slightly the long way around.

Then there are proper-class many possible values for the parameter. --Trovatore (talk) 17:43, 7 February 2011 (UTC)[reply]

Algorithmic complexity[edit]

Are weakly polynomial algorithms in P? —Preceding unsigned comment added by 207.210.136.252 (talk) 20:50, 5 February 2011 (UTC)[reply]

Yes, at least according to how "weakly polynomial" is defined in our Time complexity article. –Henning Makholm (talk) 21:09, 5 February 2011 (UTC)[reply]

Strictly speaking, P is a class of problems, not algorithms. —Bkell (talk) 04:27, 6 February 2011 (UTC)[reply]

Seemingly simple problem[edit]

A flag is going up a pole at 1 m/s. What is the rate of increase of the distance from the flag to an observer 3m from the base of the pole at time t?

This is a problem from my computer science class, and we're supposed to solve it both numerically, with a simulation, and mathematically, to demonstrate the accumulation of errors over time. The numerical part is easy, but how am I supposed to solve it mathematically? --99.237.234.245 (talk) 23:20, 5 February 2011 (UTC)[reply]

Hint: Pythagoras. —Preceding unsigned comment added by 68.9.151.24 (talk) 23:32, 5 February 2011 (UTC)[reply]

Hint 2: d/dt (Pythagoras). 71.141.88.54 (talk) 23:43, 5 February 2011 (UTC)[reply]

If the OP can do a numerical solution, he probably doesn't need to be reminded about Pythagoras. But he'll want to refresh his memory about derivatives. –Henning Makholm (talk) 23:48, 5 February 2011 (UTC)[reply]

The article related rates solves a very similar problem. Sławomir Biały (talk) 23:54, 5 February 2011 (UTC)[reply]

OK, so this requires calculus. I was kind of hoping for a geometrical solution. The problem is, I'm a grade 9 student listening in on a university computer science class, so while the professor can probably assume his students know calculus, that doesn't apply to me.

To Slavomir: using the related rates article, I decided to take a stab at the problem. According to the article, x*dx/dt + y*dy/dt = h*dh/dt. I want dh/dt, and I'm guessing dx/dt=0 while dy/dt=1 m/s. Can I plug in h=sqrt(x^2+y^2) to get dh/dt=(y*dy/dt)/sqrt(x^2+y^2)? --99.237.234.245 (talk) 00:53, 6 February 2011 (UTC)[reply]

Yes, that's the right formula.

However, let me suggest that you devote some time and energy to learning some calculus before you break your neck getting too far ahead on computer science. I don't know what kind of CS course you're taking, but if it's serious numerical methods stuff, you will need calculus as a background skill. On the other hand, if it's a generic CS 101 introductory programming thing, you can probably learn that just as well or better toying around with a computer and a programming language on your own, especially if you're bright enough to be hanging around universities in grade 9. –Henning Makholm (talk) 01:20, 6 February 2011 (UTC)[reply]

Thanks. The course is supposed to be an introduction to algorithms and algorithm analysis, with only a chapter or so on numerical methods. As far as I can tell, the math isn't difficult: "step A takes log(n) computations, but this is repeated n^2 times, so running time is some constant times n^2*log(n)". --99.237.234.245 (talk) 05:51, 6 February 2011 (UTC)[reply]

Okay. My advice still stands, speaking as a computer scientist. Algorithmics is an excellent touchstone for mathematical aptitude, so if you can wrap your head around that, learning some basic calculus by self-study would likely take you very little effort compared to the problem-solving power it will bring you. Good luck! –Henning Makholm (talk) 09:20, 6 February 2011 (UTC)[reply]